The rate law for one of the mechanism of the pyrolysis of CH3 CHO at 520 C and 0.2 bar is Rate =- |k2 (k12 k1 )1 / 2 | [CH3 CHO ]3 / 2 The overall activation energy E, in terms of the rate law is:

ABC26PH0886 · Chemical Kinetics

Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Arrhenius Activation · Exam: NET JUNE 2012 · Marks: 2 · Difficulty: Hard

The rate law for one of the mechanism of the pyrolysis of $\mathrm{CH}_{3} \mathrm{CHO}$ at $520^{\circ} \mathrm{C}$ and 0.2 bar is

\[ \text { Rate }=-\left|\mathrm{k}_{2}\left(\frac{\mathrm{k}_{1}}{2 \mathrm{k}_{1}}\right)^{1 / 2}\right|\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{3 / 2} \]

The overall activation energy E, in terms of the rate law is:

(a)$\mathrm{Ea}(2)+\mathrm{Ea}(1)+2 \mathrm{Ea}(4)$

(b)$\mathrm{Ea}(2)+\frac{1}{2} \mathrm{Ea}(1)-\mathrm{Ea}(4)$

(c)$\mathrm{Ea}(2)+\frac{1}{2} \mathrm{Ea}(1)-\frac{1}{2} \mathrm{Ea}(4)$

(d)$\mathrm{Ea}(2)-\frac{1}{2} \mathrm{Ea}(1)-\frac{1}{2} \mathrm{Ea}(4)$

Answer

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